A ray of light of intensity $I$ is incident on a parallel glass-slab at a point $A$ as shown in fig. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays $AB$ and $A'B'$ undergo interference. The ratio $I_{\max}/I_{\min}$ is

Question

Dr. Abubacker · Accepted Answer

(d) From figure $I_{1} = \frac{I}{4}$ and $I_{2} = \frac{9I}{64} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{9}{16}$
By using $\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{\frac{I_{2}}{I_{1}}} + 1}{\sqrt{\frac{I_{2}}{I_{1}}} - 1} \right)^{2} = \left( \frac{\sqrt{\frac{9}{16}} + 1}{\sqrt{\frac{9}{16}} - 1} \right)^{2} = \frac{49}{1}$

Q. A ray of light of intensity $I$ is incident on a parallel glass-slab at a point $A$ as shown in fig. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays $AB$ and $A'B'$ undergo interference. The ratio $I_{\max}/I_{\min}$ is

Explanation

Reviewed by Dr. Abubacker

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